It can be shown that $( 1 + x ) ^ { n } = 1 + n x + \frac { n ( n - 1 ) } { 2 ! } x ^ { 2 } + \frac { n ( n - 1 ) ( n - 2 ) } { 3 ! } x ^ { 3 } \ldots$ is true for any real number n (not just positive
integer values) and any real number x, where $| x |

Question

It can be shown that $( 1 + x ) ^ { n } = 1 + n x + \frac { n ( n - 1 ) } { 2 ! } x ^ { 2 } + \frac { n ( n - 1 ) ( n - 2 ) } { 3 ! } x ^ { 3 } \ldots$ is true for any real number n (not just positive
integer values) and any real number x, where $| x | < 1$ Use this series to approximate the given number to the nearest
thousandth.
-$1.064$

Quizplus · Accepted Answer

The Answer of It can be shown that $( 1 + x ) ^ { n } = 1 + n x + \frac { n ( n - 1 ) } { 2 ! } x ^ { 2 } + \frac { n ( n - 1 ) ( n - 2 ) } { 3 ! } x ^ { 3 } \ldots$ is true for any real number n (not just positive
integer values) and any real number x, where $| x | < 1$ Use this series to approximate the given number to the nearest
thousandth.
-$1.064$

It Can Be Shown That (1+x)n=1+nx+n(n−1)2!x2+n(n−1)(n−2)3!x3…( 1 + x ) ^ { n } = 1 + n x + \frac { n ( n - 1 ) } { 2 ! } x ^ { 2 } + \frac { n ( n - 1 ) ( n - 2 ) } { 3 ! } x ^ { 3 } \ldots(1+x)n=1+nx+2!n(n−1)​x2+3!n(n−1)(n−2)​x3…

It Can Be Shown That $( 1 + x ) ^ { n } = 1 + n x + \frac { n ( n - 1 ) } { 2 ! } x ^ { 2 } + \frac { n ( n - 1 ) ( n - 2 ) } { 3 ! } x ^ { 3 } \ldots$